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INTRODUCTION
In Sawfy, the picaro encloses a space and creates a black hole in it. I will be calculating the energy required to create this black hole.
CALCULATION
I will be taking the average height of the piccaro, it is stated in the story that some of the piccaro are shorter while others were taller so i will be taking the average of height of the picaro being equal to the height of yachiru since they are implied to be relative size in the story and the official wiki height of yachiru 3ft.7 inch = 1.09m
Average picaro height = 1.09m
Since the number of picaro present at the feat was close to hundred at the time of the feat and I will be using hundred as default (circumference lowball) and assume that no spacing existed within them which is consistent about the picaro enclosing the space around allowing no energy to escape.
Circumference of picaro enclosure = 100 × 0.273m= 27.3m
Diameter and radius of the circle
Radius= 27.3/2π = 4.35m
Diameter= 2r= 8.70m
Assuming that the picaro would individually from their enclosed position (circumference) is looking a VISIBLE black sphere at 4.35m away at the radius from their position thus;
Visual angular resolution for a normally sighted human (20/20 vision) is about 1 arcminute (1′) ≈ π/10,800 ≈ 0.000290888 rad.
Linear size s subtending that angle at distance ddd is roughly
s= d × (tan(θ)) = d × (θ) for very small (θ)
Use d=4.35m
Results (smallest resolvable linear size at 4.35 m)
20/20 vision (1′):
θ=1′ → s = 4.35×0.000290888=0.001265 m = 1.27mm
Since the picaro are in the sky at the time of the feat thus they should have good lightning to see considerably at the distance to aleast that of a normal human acuity resolution.
The diameter of the black hole from the perspective they were all at Since it was atleast visible is =
1.27mm
Radius of black hole = 0.00127/2 = 0.000635 m
"Also to point out, any black hole within up to Low 4-C should be calculated with this formula: ((299792458^2/(2*6.674*10^-11))/(5.972*10^24))*(2.487*10^32) * radius in meters = energy in joules"
Using vswiki standards:
299792458^2/ (2*6.674*10^-11)/(5.972*10^24))&(2.487*10^32)* 0.000635 =
1.78055305721505*10^31 joules = 4.257 Zetta tons [Small Planet Level]
Please I need your thoughts on this for any correction
In Sawfy, the picaro encloses a space and creates a black hole in it. I will be calculating the energy required to create this black hole.
CALCULATION
I will be taking the average height of the piccaro, it is stated in the story that some of the piccaro are shorter while others were taller so i will be taking the average of height of the picaro being equal to the height of yachiru since they are implied to be relative size in the story and the official wiki height of yachiru 3ft.7 inch = 1.09m
Average picaro height = 1.09m
Since the number of picaro present at the feat was close to hundred at the time of the feat and I will be using hundred as default (circumference lowball) and assume that no spacing existed within them which is consistent about the picaro enclosing the space around allowing no energy to escape.
Circumference of picaro enclosure = 100 × 0.273m= 27.3m
Diameter and radius of the circle
Radius= 27.3/2π = 4.35m
Diameter= 2r= 8.70m
Assuming that the picaro would individually from their enclosed position (circumference) is looking a VISIBLE black sphere at 4.35m away at the radius from their position thus;
Visual angular resolution for a normally sighted human (20/20 vision) is about 1 arcminute (1′) ≈ π/10,800 ≈ 0.000290888 rad.
Linear size s subtending that angle at distance ddd is roughly
s= d × (tan(θ)) = d × (θ) for very small (θ)
Use d=4.35m
Results (smallest resolvable linear size at 4.35 m)
20/20 vision (1′):
θ=1′ → s = 4.35×0.000290888=0.001265 m = 1.27mm
Since the picaro are in the sky at the time of the feat thus they should have good lightning to see considerably at the distance to aleast that of a normal human acuity resolution.
The diameter of the black hole from the perspective they were all at Since it was atleast visible is =
1.27mm
Radius of black hole = 0.00127/2 = 0.000635 m
"Also to point out, any black hole within up to Low 4-C should be calculated with this formula: ((299792458^2/(2*6.674*10^-11))/(5.972*10^24))*(2.487*10^32) * radius in meters = energy in joules"
Using vswiki standards:
299792458^2/ (2*6.674*10^-11)/(5.972*10^24))&(2.487*10^32)* 0.000635 =
1.78055305721505*10^31 joules = 4.257 Zetta tons [Small Planet Level]
Please I need your thoughts on this for any correction
Last edited: